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a) \(\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}+\frac{4x}{a+b+c}=1\)
\(\Leftrightarrow\frac{a+b-x}{c}+1+\frac{b+c-x}{a}+1+\frac{c+a-x}{b}+1+\frac{4x}{a+b+c}-4=0\)
\(\Leftrightarrow\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}+\frac{4x-4\left(a+b+c\right)}{a+b+c}=0\)
\(\Leftrightarrow\left(x-a-b-x\right)\left(\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}\right)=0\)
b)đề bài như trên
\(\Leftrightarrow\left(\frac{x-a-b-c}{bc}\right)+\left(\frac{x-b}{ca}-\frac{1}{a}-\frac{1}{c}\right)+\left(\frac{x-c}{ab}-\frac{1}{a}-\frac{1}{b}\right)=0\)
\(\Leftrightarrow\left(x-a-b-c\right)\left(\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}\right)=0\)
\(\frac{x-a}{bc}+\frac{x-b}{ac}+\frac{x-c}{ab}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\frac{x-a}{bc}+\frac{x-b}{ac}+\frac{x-c}{ab}=\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\)
\(\frac{ax-a^2+bx-b^2+cx-c^2}{abc}=2\left(\frac{ab+bc+ac}{abc}\right)\)
\(ax-a^2+bx-b^2+cx-c^2=2\left(ab+bc+ac\right)\)
\(x\left(a+b+c\right)-\left(a^2+b^2+c^2\right)=2\left(ab+bc+ac\right)\)
\(x\left(a+b+c\right)=a^2+b^2+c^2+2ab+2bc+2ac\)
\(x=a+b+c\)
\(c,\frac{x-a-b}{c}-1+\frac{x-b-c}{a}-1+\frac{x-a-c}{b}-1=0.\)
\(\frac{x-a-b-c}{c}+\frac{x-a-b-c}{a}+\frac{x-a-b-c}{b}=0\)
\(\left(x-a-b-c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\)
=>\(\orbr{\begin{cases}a+b+c=x\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\end{cases}}\)
Vậy.......
Ta có \(\frac{x-b-c}{a}+\frac{x-c-a}{b}+\frac{x-b-a}{c}=3\)
\(\Rightarrow\frac{x-b-c}{a}+\frac{x-c-a}{b}+\frac{x-b-a}{c}-3=0\)
\(\Leftrightarrow\left(\frac{x-b-c}{a}-1\right)+\left(\frac{x-c-a}{b}-1\right)+\left(\frac{x-b-a}{c}-1\right)=0\)
\(\Leftrightarrow\frac{x-a-b-c}{a}+\frac{x-a-b-c}{b}+\frac{x-a-b-c}{c}=0\)
\(\Leftrightarrow\left(x-a-b-c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\)
Vì \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ne0\) nên chỉ có
x-a-b-c=0 =>x=a+b+c
Vậy x=a+b+c
1)\(\frac{x-b-c}{a}+\frac{x-c-a}{b}+\frac{x-a-b}{c}=3\)
=>\(\frac{x-b-c}{a}-1+\frac{x-c-a}{b}-1+\frac{x-a-b}{c}-1=0\)
=>\(\frac{x-a-b-c}{a}+\frac{x-a-b-c}{b}+\frac{x-a-b-c}{c}=0\)
=>\(\left(x-a-b-c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\)
Nếu x - a -b -c = 0 => phương trình có nghiệm duy nhất x = a + b + c
Nếu \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)=> Phương trình có vô số nghiệm x thuộc R
\(ĐKXĐ:a,b,c\ne0\)
\(\frac{x-a}{bc}+\frac{x-b}{ca}+\frac{x-c}{ab}=\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\)
\(\Leftrightarrow\frac{xa-a^2}{abc}+\frac{xb-b^2}{abc}+\frac{xc-c^2}{abc}=\frac{2bc}{abc}+\frac{2ac}{abc}+\frac{2ab}{abc}\)
\(\Leftrightarrow\frac{xa-a^2+xb-b^2+xc-c^2}{abc}=\frac{2bc+2ac+2ab}{abc}\)
\(\Leftrightarrow xa-a^2+xb-b^2+xc-c^2=2bc+2ac+2ab\)
\(\Leftrightarrow xa+xb+xc=2bc+2ac+2ab+a^2+b^2+c^2\)
\(\Leftrightarrow x\left(a+b+c\right)=\left(a+b+c\right)^2\)
\(\Leftrightarrow x=a+b+c\)
Vậy x = a + b + c
\(ĐKXĐ:a,b,c\ne0\)
\(\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}+\frac{4x}{a+b+c}=1\)
\(\Leftrightarrow\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}=1-\frac{4x}{a+b+c}\)
\(\Leftrightarrow1+\frac{a+b-x}{c}+1+\frac{b+c-x}{a}+1+\frac{c+a-x}{b}=4\)
\(-\frac{4x}{a+b+c}\)
\(\Leftrightarrow\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}=\)
\(\frac{4\left(a+b+c\right)}{a+b+c}-\frac{4x}{a+b+c}\)
\(\Leftrightarrow\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}=\)
\(\frac{4\left(a+b+c-x\right)}{a+b+c}\)
\(\Leftrightarrow\left(a+b+c-x\right)\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}-\frac{4}{a+b+c}\right)=0\)
\(\Rightarrow\left(a+b+c-x\right)=0\)hoặc \(\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}-\frac{4}{a+b+c}\right)=0\)
+) Nếu \(\Rightarrow\left(a+b+c-x\right)=0\)thì x = a + b + c
+) Nếu \(\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}-\frac{4}{a+b+c}\right)=0\)thì x thỏa mãn với mọi số